Divide the following complex numbers. $ \dfrac{-3-5i}{-5+3i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-5-3i}$ $ \dfrac{-3-5i}{-5+3i} = \dfrac{-3-5i}{-5+3i} \cdot \dfrac{{-5-3i}}{{-5-3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-3-5i) \cdot (-5-3i)} {(-5+3i) \cdot (-5-3i)} = \dfrac{(-3-5i) \cdot (-5-3i)} {(-5)^2 - (3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-3-5i) \cdot (-5-3i)} {(-5)^2 - (3i)^2} = $ $ \dfrac{(-3-5i) \cdot (-5-3i)} {25 + 9} = $ $ \dfrac{(-3-5i) \cdot (-5-3i)} {34} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-3-5i}) \cdot ({-5-3i})} {34} = $ $ \dfrac{{-3} \cdot {(-5)} + {-5} \cdot {(-5) i} + {-3} \cdot {-3 i} + {-5} \cdot {-3 i^2}} {34} $ Evaluate each product of two numbers. $ \dfrac{15 + 25i + 9i + 15 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{15 + 25i + 9i - 15} {34} = \dfrac{0 + 34i} {34} = i $